If P = $\begin{bmatrix$}[r] 2 & 6 \ 3 & 9 \end{bmatrix} \text{ and Q } = \begin{bmatrix}[r] 3 & x \ y & 2 \end{bmatrix}, find x, y such that PQ = O.

Given,

PQ = 0

$\Rightarrow \begin{bmatrix$}[r] 2 & 6 \ 3 & 9 \end{bmatrix} \begin{bmatrix}[r] 3 & x \ y & 2 \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times 3 + 6 \times y & 2 \times x + 6 \times 2 \ 3 \times 3 + 9 \times y & 3 \times x + 9 \times 2 \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6 + 6y & 2x + 12 \ 9 + 9y & 3x + 18 \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

⇒ 6 + 6y = 0 and 2x + 12 = 0
⇒ 6y = -6 and 2x = -12
⇒ y = -1 and x = -6.

Checking whether x = -6 and y = -1, satisfies other equations 9 + 9y = 0 and 3x + 18 = 0,

⇒ 9 + 9y = 0
⇒ 9 + 9(-1) = 0
⇒ 9 - 9 = 0 (L.H.S. = R.H.S.)

⇒ 3x + 18 = 0
⇒ 3(-6) + 18 = 0
⇒ -18 + 18 = 0 (L.H.S. = R.H.S.)

∴ x = -6 and y = -1.

Hence, the value of x = -6 and y = -1.